Problem: Subtract the following rational expressions. $\dfrac{8k}{9k-4}-\dfrac{3k^3}{2k+7}=$
Answer: We can subtract two rational expressions whose denominators are equal by subtracting the numerators and keeping the denominator the same. [Does this fit with how we subtract rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators: $({9k-4})\cdot({2k+7})$. Let's manipulate the expressions to have that denominator: $\begin{aligned} &\phantom{=}\dfrac{8k}{{9k-4}}-\dfrac{3k^3}{{2k+7}} \\\\ &=\dfrac{8k\cdot({2k+7})}{({9k-4})\cdot({2k+7})}-\dfrac{3k^3\cdot({9k-4})}{({2k+7})\cdot({9k-4})} \end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's subtract! $\begin{aligned} &\phantom{=}\dfrac{8k\cdot(2k+7)}{(9k-4)\cdot(2k+7)}-\dfrac{3k^3\cdot(9k-4)}{(2k+7)\cdot(9k-4)} \\\\ &=\dfrac{8k\cdot(2k+7)-3k^3\cdot(9k-4)}{(9k-4)(2k+7)} \\\\ &=\dfrac{16k^2+56k-27k^4+12k^3}{(9k-4)(2k+7)} \\\\ &=\dfrac{-27k^4+12k^3+16k^2+56k}{(9k-4)(2k+7)} \end{aligned}$ In conclusion, $\dfrac{8k}{9k-4}-\dfrac{3k^3}{2k+7}=\dfrac{-27k^4+12k^3+16k^2+56k}{(9k-4)(2k+7)}$